6.

Let x be the independent variable. We have

$$\frac{dv}{dy} = \frac{dv/dx}{dy/dx} = \frac{12 y^2}{v}$$

This is separable, and the solution is $$v^2 = 8 y^3 + C$$. From the initial conitions, C = 0. Now $$y' = v = (2 y)^{3/2}$$ (with a + sign since v(0) is positive). This equation is separable, and the general solution is $$-2 y^{-1/2} = 2^{3/2} x + K$$. Again using the initial conditions, $$K = - 2^{1/2}$$. The solution can be written as y = 2/(2 x - 1)². It is defined for $-\infty < x < 1/2$.